Question

what is the difference of degrees centigrade and centigrade degrees?

Answer 1

My teacher said centigrade degrees is the change in temperature of degrees centigrade and it is also equal to kelvin.
is this correct?
if yes, why did the change in degrees centigrade became kelvin?

Answer 2

I've never heard of that distinction, but the difference between two temperature mesured in centigrade is equivalent to its change in temperature in Kelvin.

\[0ºC = 273.15 K\]
\[36ºC = (36 + 273.15) K\]

So if you have a change in temperature measured in centigrade

\[45ºC - 12ºC = 33ºC
\\ \hspace{80px}= (45 + 273.15)K - (12+273.15)K
\\ \hspace{80px}= \big(45-12 - (273.15-273.15)\big)K
\\ \hspace{80px}= (45-12)K
\\ \hspace{80px}=33K\]

Answer 3

Thanks! So the change in centigrade can be written as kelvin or centigrade because the constant just cancels out when you convert it to kelvin then subtract.
am I right?

Answer 4

yup yup ^_^

Answer 5

do you know what rankine is?

Answer 6

I had not heard of it until just now - it's another temperature scale?

Answer 7

yes

Answer 8

ooo, it's just a constant added to Fahrenheit instead of a constant added to Celsius. I see

Answer 9

It's an old temperature scale that nobody uses.
iirc, it's like Kelvin, but for Fahrenheit.

Answer 10

my teacher listed it on the temperature scale. I don't what it is

Answer 11

What @Lessis said

Answer 12

Basically, Rankin is just Fahrenheit transformed into an absolute scale of temperature. Still, nobody uses it (It's not relatable to the SI units, unless you transform it to Kelvin).

Answer 13

so its just a fancy term for directly converting Fahrenheit to kelvin? because you cannot convert Fahrenheit to kelvin directly.

Answer 14

No. It's a whole different scale than Kelvin. The only thing they have in common is that 0 Kelvin is the same as 0 Rankin.

Answer 15

http://en.wikipedia.org/wiki/Rankine_scale
The table at the bottom tells you how to relate some important temperatures in different scales. (Also, conversion formulae)

Answer 16

I understand it now.
Thanks to both of you for clarification.
:)

Answer 17

^_^ welcome

Answer 18

No problem. :D